Problem: If $f'(x)=[f(x)]^2$ and $f(0)=1$, then $f(6)=1/n$ for some integer $n$. What is $n$ ? $n=~$
Explanation: If we let $y=f(x)$ and change the derivative notation, we can rewrite the differential equation so that it's separable. $\dfrac{dy}{dx}=y^2$ What does it look like after we separate the variables? $y^{-2}\,dy=dx$ Let's integrate both sides of the equation. $\int y^{-2}\,dy=\int dx$ What do we get? $-y^{-1}=x+C$ What value of $C$ satisfies the initial condition $y(0)=1$ ? Let's substitute $x=0$ and $y=1$ into the equation and solve for $C$. $\begin{aligned} -1^{-1}&=0+C\\ \\ C&=-1 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} -y^{-1}&=x-1\\ \\ y^{-1}&=1-x\\ \\ y&=\dfrac1{1-x} \end{aligned}$ If $y(6)=1/n$, what is the integer $n$ ? $\begin{aligned} y(6)&=\dfrac1{1-6}\\ \\ \\ &=\dfrac1{-5}\\ \\ \\ &=\dfrac1n \end{aligned}$ Thus, $n=-5$.